Symmetric Equation of Line in 3D converts to 2 Planes or 2 Lines?

[Ocata]

Homework Statement

Hi,

An equation of the form Ax + By + C = 0 is a standard equation of a line in 2D.

An equation of the form Ax + By + Cz + D = 0 is an equation of a plane.

Is it possible to:

Describe a plane in space, written in standard form, such that one variable is missing from the equation?
For instance, if z = 0, then Ax + By + Cz + D = 0 becomes Ax+By+D=0.

If z = 0, then the equation of the plane can no longer describe a plane, but instead only describe a line, correct?

Reason for the question is because when a vector equation of a line described by two points, say (1,2,3) and (5,7,10), a vector equation in symmetric form can be written as such:

$$\frac{x-1}{4}= \frac{y-2}{5} = \frac{z-3}{7}$$

which simplifies to two equations which seem to describe two lines instead of two planes:

$\frac{x-1}{4}= \frac{y-2}{5} = \frac{z-3}{7}$

$\frac{x-1}{4} =\frac{y-2}{5}$ and $\frac{y-2}{5}= \frac{z-3}{7}$


$-5 \big(x-1)+4 \big(y-2) = 0$ and $7 \big(y-2) -5 \big(z-3) =0$

$-5x + 4y - 3 = 0$ and $7y-5z+1 = 0$

As can be seen, these are two equations of two variables. I understand this to describe two lines. One line in the xy plane and another line in the yz plane. It doesn't make sense to me that two lines intersect to make a new line. So how is it that these two seemingly linear equations are actually two equations of planes? Is it possible that I just don't understand the equation of a plane well enough?

Homework Equations

$-5x + 4y - 3 = 0$ and $7y-5z+1 = 0$

The Attempt at a Solution

Guess #1: would be that these two equations are in fact linear equations, represent 2 lines instead of 2 planes, and I am calculating wrong to arrive a such equations.

Guess #2: would be that these two equations are actually equations of planes and due to 3 variables being utilized across 2 equations (even if each equation only contains 2 variables) some how represents two planes.

 

[LCKurtz]

Homework Statement

Hi,

An equation of the form Ax + By + C = 0 is a standard equation of a line in 2D.

An equation of the form Ax + By + Cz + D = 0 is an equation of a plane.

Is it possible to:

Describe a plane in space, written in standard form, such that one variable is missing from the equation?
For instance, if z = 0, then Ax + By + Cz + D = 0 becomes Ax+By+D=0.

If z = 0, then the equation of the plane can no longer describe a plane, but instead only describe a line, correct?

No. If a variable is missing, that just means the plane is parallel to the axis of the missing variable.

Reason for the question is because when a vector equation of a line described by two points, say (1,2,3) and (5,7,10), a vector equation in symmetric form can be written as such:

$$\frac{x-1}{4}= \frac{y-2}{5} = \frac{z-3}{7}$$

which simplifies to two equations which seem to describe two lines instead of two planes:

$\frac{x-1}{4}= \frac{y-2}{5} = \frac{z-3}{7}$

$\frac{x-1}{4} =\frac{y-2}{5}$ and $\frac{y-2}{5}= \frac{z-3}{7}$


$-5 \big(x-1)+4 \big(y-2) = 0$ and $7 \big(y-2) -5 \big(z-3) =0$

$-5x + 4y - 3 = 0$ and $7y-5z+1 = 0$

As can be seen, these are two equations of two variables. I understand this to describe two lines. One line in the xy plane and another line in the yz plane. It doesn't make sense to me that two lines intersect to make a new line. So how is it that these two seemingly linear equations are actually two equations of planes? Is it possible that I just don't understand the equation of a plane well enough?

Yes, it is possible you need to study it a bit more. When you take any pair of the symmetric equalities they represent two planes and a point (x,y,z) is on both only if it is on their line of intersection. So the pair of equations, which are themselves representing planes, together represent their line of intersection.